Probability: How two gentlemen argued over an unfinished game

Question

How do you divide the stake from a game that was never finished?

The interrupted match

Two players, A and B, play a series of games. Each game ends with a win for one of them; there are no draws, neither has the upper hand, and the result of one game says nothing about the next. They have put a stake into the pot, and whoever first wins six games takes all of it. But the match is interrupted when A leads 5 : 3. The stake lies on the table. Who gets it, and how much?

The problem had been known in Italy for more than a century, and everyone solved it differently. Listen to how they might have quarrelled about it, had they been able to write to each other.

A chat about the unfinished game

Four Italian scholars argue over the stake. At the end you choose whom to ask first: Pascal or Fermat.

Chat: “The unfinished game”

an imagined conversation

The messages have not survived; the ideas have. A match to six wins was interrupted at 5 : 3 to player A.

Luca PacioliLet us divide by the points won. A has five, B three, so we split the stake 5 : 3. Simple.

Niccolò TartagliaThat doesn’t sit right with me. I take the points needed to win as the whole, and the points won are a part of it. From those ratios I then work out the opponent’s share too.

Lorenzo ForestaniI would split the uncertain remainder in half. We don’t know who will win, so let both have the same chances.

Girolamo CardanoYou are all looking backwards, at what has happened. Look forwards: how could the game end? Hypothetical scores!

Luca PacioliAnd how will you count them?

Girolamo CardanoI don’t know yet. I lack the words for it.

None of them can prove they are right. What is missing is a language in which such a question could be discussed precisely. In 1654 two gentlemen from Paris and Toulouse join the conversation. Whom will you ask first?

Who’s who – four voices before 1654

Luca Pacioli, friar and teacher of reckoning: Summa de arithmetica (1494). Pictured with a pupil at a table.

Girolamo Cardano, physician and mathematician: Practica arithmeticae (1539).

Niccolò Tartaglia, mathematician from Brescia: General trattato di numeri et misure (1556).

Lorenzo Forestani, Franciscan: Pratica d’arithmetica e geometria (1603). We don’t know what he looked like, so he is shown by the title page of a later edition.

Both answers arrive at the same number, and today we would say it is about the chances of winning. But chance in today’s sense did not exist back then. An unfinished game is only a hypothetical scenario, something that did not happen, and asking what might hypothetically have happened made no sense. What happens was decided by God, not by a count of cases. So people did not even ask about chance. And what is chance, anyway?

How did such a question reach Pascal? Picture two men at a table, playing for gold pistoles (the pistole was a gold coin then current in France and Spain). One is winning, the other losing. In comes a messenger: the king is calling the knight to battle, and the game must stop. The one who was ahead is indignant that victory is slipping through his fingers. The other, who was losing, is not troubled. After all, his opponent has no idea how to divide the stake fairly. And perhaps nobody does.

It was just such a question that the chevalier de Méré, a courtier and passionate gambler, put to Pascal in 1654. Who Pascal and Fermat were, and how they dealt with it, is the subject of the following parts.

Person

A Parisian mathematician and a lawyer from Toulouse who wrote each other letters.

Two correspondents

The story has two main characters, and both came to it from somewhere you would not expect. Pascal was a philosopher and inventor, Fermat a lawyer who did mathematics in his spare time. Together they created a field in a single exchange of letters. Whether they ever met in person is not certain.

Blaise Pascal

1623–1662, Clermont → Paris → Rouen

Mathematician, physicist, philosopher and moralist. In Paris he lived “freely”, among salons and gambling. That is where he met Méré.

1623Born on 19 June in Clermont.
1640Essay pour les coniques: projective geometry.
1642The Pascaline, a mechanical calculator.
1654In summer, letters with Fermat. In autumn he escapes death in a carriage accident, and on 23 November he lives through a mystical night. He joins the Jansenists at Port-Royal.
1662Dies in Paris on 19 August.
1670Les Pensées is published posthumously.

The Pascaline – a machine that calculates

A mechanical calculator made of cogwheels, which Pascal built as a young man. Napier wanted to save people from multiplying; Pascal handed the work to machines. Ten years later he was investigating how to calculate with things that have not happened yet.

Pierre de Fermat

1601–1665, Beaumont-de-Lomagne → Toulouse

Lawyer and councillor of the Parliament of Toulouse. Mathematics was his hobby, and in it he was one of the very greatest.

1601Born in Beaumont-de-Lomagne.
1631Becomes a councillor of the Parliament in Toulouse.
1654Letters with Pascal on dividing the stake.
1660Proposes to Pascal that they meet halfway. We don’t know whether they did.
1665Dies on 12 January in Castres.

To speak frankly to you about geometry, I find it the highest exercise of the mind …

Pour vous parler franchement de la Géométrie, je la trouve le plus haut exercice de l’esprit …

… and so I call it the finest craft in the world …

… aussi je l’appelle le plus beau métier du monde …

… the one in all of Europe whom I hold to be the greatest geometer …

… celui de toute l’Europe que je tiens pour le plus grand géomètre …

Pascal to Fermat, 10 August 1660

So Pascal wrote to Fermat six years later. He added at once that, when all is said and done, it is only a craft.

Who asked

The question was not posed by a mathematician but by a gambler. The chevalier de Méré (real name Antoine Gombaud) played dice for money and had a rule that earned him money for a long time: bet that in four throws of a single die at least one six will come up. Then he changed the game. He started throwing two dice and betting that in 24 throws a double six would come up at least once.

The reasoning sounded sensible. Two dice have 36 possible pairs instead of six, six times as many. If four throws were enough, six times as many throws should do, that is 4 · 6 = 24. But at the table it did not work out, and he had no idea why. With this doubt he turned to Pascal.

Where Méré went wrong – why 24 throws are not enough

Chances do not add, they multiply. It is easier to count the opposite: that a double six never comes up. In one throw of two dice this happens in 35 cases out of 36, so with a chance of 35/36.

In 24 throws it has to fail all 24 times in a row: (35/36)²⁴ ≈ 0.509. That is more than half, so a double six comes up in only about 49.1 % of cases. Méré was betting against himself. With 25 throws it is (35/36)²⁵ ≈ 0.494, a chance of about 50.6 %. The smallest number of throws for which the bet pays is 25.

With one die and four throws it works out correctly: (5/6)⁴ ≈ 0.482, so the chance of at least one six is about 51.8 %. That is why his original rule worked. But the proportion “six times as many possibilities, six times as many throws” does not hold for two dice.

Idea

Three letters, two paths, one result.

Looking forward

Cardano already held the key: look not at what has happened, but at what may still come. But nobody knew what to do with it. Pascal and Fermat finished the game, on paper only, each in his own way. Their dispute and their agreement played out in the letters you see below.

The letters are rolled up. Click any of them and it will unfold.

I see that the truth is the same in Toulouse as in Paris.

Je vois bien que la vérité est la même à Toulouse et à Paris.

Pascal begins joyfully: he and Fermat agree on the dice problem. Then he puts another problem to him. Two players play for three winning games, each has staked 32 pistoles (gold coins), 64 in total. The game is interrupted. Depending on how far they got, one of three cases arises.

Game to three winseach staked 32 pistoles, 64 in total
- 1.2 : 1Pascal needs 1 more win, Fermat 248 : 16
- 2.2 : 0Pascal needs 1 more win, Fermat 356 : 8
- 3.1 : 0Pascal needs 2 more wins, Fermat 344 : 20

Here Pascal and Fermat play the two players themselves; in the letters the players have no names. Each scenario builds on the result of the previous one.

1. Scenario 2 : 1

Score 2 : 1next game
- ½Bingo! Takes all 64
- ½score 2 : 2, even chances
  - ½takes 64
  - ½takes 64

If Pascal wins, it is over and he takes all 64. If Fermat wins, it is 2 : 2 and fortune is even. So Pascal already holds 32 pistoles for certain, and the remaining 32 are played for at even odds.

Chances ¾ : ¼, that is 48 : 16

2. Scenario 2 : 0

Score 2 : 0next game
- ½takes all 64
- ½score 2 : 1, see scenario 1: Pascal holds 48

Pascal has 48 for certain, because even if he lost the next game he would end up in scenario 2 : 1. For the remaining 16, fortune is even.

56 : 8, because ½ · 64 + ½ · 48 = 56

3. Scenario 1 : 0

Score 1 : 0next game
- ½score 2 : 0, see scenario 2: Pascal holds 56
- ½score 1 : 1, even chances: Pascal holds 32

Pascal has 32 for certain. The difference 56 − 32 = 24 is played for at even odds, which makes 12.

44 : 20, because ½ · 56 + ½ · 32 = 44

In the letter Pascal sums up how Fermat proceeds: count the greatest possible number of games that would have to be played. Player A needs 2 more wins and player B 3, so the match ends after four games at the latest. “We throw a die with two faces.” Four throws give 2⁴ = 16 possible endings. All that remains is to count them.

4 games2⁴ = 16 possible endings
- AA wins11 of 16 endings
  - 1A wins 4 of 4
  - 4A wins 3 of 4
  - 6A wins 2 of 4
- BB wins5 of 16 endings
  - 4B wins 3 of 4
  - 1B wins 4 of 4

Division 11 : 5, with a stake of 64 pistoles 44 : 20

The result is the same as in scenario 1 : 0 from the first letter. A player who needs two wins against three is owed the same share of the stake whether you count it by steps backwards or by all possible endings.

Sabotage: Monsieur Roberval’s objection

Roberval: There is no need to play four games. The match may end sooner.

Pascal: Nothing prevents carrying on. If games are played that no longer decide anything, the score holds “unchanged” and it has no effect on dividing the stake. The extra games are just a bookkeeping device for counting all endings in the same way.

… if I am not mistaken, such a division is wrong …

… si je ne me trompe, ce parti est mal juste …

In the same letter, though, Pascal hit a snag. With three players he was not sure the method gave the right division. Throwing a three-faced die without keeping track of the order of the results leads to a wrong count. The throws have to be taken one after another.

Fermat replies, and he has a single answer: order. Players A, B and C need 1, 2 and 2 wins respectively. The match ends after three games at the latest, and any of them can win each game. The order of three results gives 3 · 3 · 3 = 27 possible endings.

3 games, 3 players3³ = 27 orders of results
- AA wins17 of 2717
- BB winsBBA, BBB, BBC, BCB, CBB5
- CC winsCCA, CCC, CCB, CBC, BCC5

Division 17 : 5 : 5

What is new here is precisely the model. A two- or three-faced “die” is thrown repeatedly and every order of results is counted separately. Then dividing the stake is just counting the favourable orders.

The Triangle arithmétique, Pascal’s triangle, also appears in the correspondence of the same period. It shortens counting sixteen, twenty-seven or more endings to adding up the numbers in one row. How, the next part will show.

Pascal worked through scenarios backwards from the end of the game, Fermat counted all possible endings. The paths were different, the number the same. Out of that agreement came something more than a table of pistoles: a new way of talking about things that have yet to happen.

Formula

Only now the notation: favourable endings over all possible ones.

What do the letters have in common?

Pistoles, players and dice are just scenery. Beneath them lies a single idea: the fair share of an uncertain win is the number of endings that lead to your win, divided by the number of all possible endings.

Pascal then found that he did not have to write these counts out. They lie in a row of his arithmetical triangle. Let n₁ and n₂ be the numbers of wins that players A and B still need, and N = n₁ + n₂. Add up the first n₂ numbers of the row with N numbers and divide by the sum of the whole row.

Divide the stake yourself

Set how many wins each player still needs. Try A: 2, B: 3 and you will get the number from the second letter.

Pascal’s triangle as a counting board

row with N numbers

Player A needs wins Player B needs wins Stake

The highlighted numbers are the endings won by player A. Their sum divided by the sum of the whole row is A’s share.

FORMULA

share = favourable endingsall possible endings

share of A = C(N−1, 0) + … + C(N−1, n₂−1)2^N−1 · S

The future, counted piece by piece.

Appendix for the curious

Pascal’s wager

The heart has its reasons, which reason does not know.

Le cœur a ses raisons que la raison ne connaît point.

Pascal, Les Pensées, 1670

After the mystical night of 23 November 1654, Pascal carried the same way of reasoning (weigh up what may come and decide accordingly) over to a question that has nothing to do with dice. Man, in his view, is like a thinking reed, suspended between the infinitely small and the infinitely large, unhappy and seeking diversion. And faith, for him, is a wager in which one has to decide.

	God is an illusion	God is real
Atheist	loses nothing, has their diversions	loses eternity, has their diversions
Believer	doesn’t know they have lost; also has diversions, but through doing good	loses nothing and gains eternity

This is what Pascal’s argument looks like at a glance. Its validity has been disputed for centuries. We include it here because it shows how far from games of chance the new way of thinking about uncertainty travelled within a single lifetime.